What is the 16th term of the arithmetic sequence whose common difference is d=9 and whose first term is a=2

Answer 1

a₁ = first term = 2

a₂ = 2nd term = a₁ + 9 = 11

a₃ = a₂+9 = 11 + 9 = 20

a₄ = 20 + 9 = 29

a₅ = 29 + 9 = 38

a₆ = 38 + 9 = 47

a₇ = 47 + 9 =56

a₈ = 56 + 9 = 65

a₉ = 65 + 9 = 74

a₁₀ = 74 + 9 = 83

a₁₁ = 83 + 9 = 92

a₁₂ = 92 + 9 = 101

a₁₃ = 1041 + 9 = 110

a₁₄ = 110 + 9 = 119

a₁₅ = 119 + 9 = 128

a₁₆ = 128 + 9 = 137

Or you could use the formula a_n = a₁ + (n-1)(d)

Where n = 16, d = 9, and a₁ = 2

Doing it this way we find that a₁₆ = 2 + (16 - 1)(9) = 2 + (15)(9) = 137

NOTICE we get the same answer (a₁₆ = 137) either way!